Question

A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Use a 0.01 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute?

 72 79 42 66 41 23 63 65 67 48 65 72 96 88 68 

Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? 

Answer 1

Solution:

x	x2
72	5184
79	6241
42	1764
66	4356
41	1681
23	529
63	3969
65	4225
67	4489
48	2304
65	4225
72	5184
96	9216
88	7744
68	4624
∑x=955	∑x2=65735

Mean ˉx=∑xn

=72+79+42+66+41+23+63+65+67+48+65+72+96+88+68/15

=955/15

=63.6667

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√65735-(955)215/14

=√65735-60801.6667/14

=√4933.3333/14

=√352.381

=18.7718

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :  μ  = 60

H_a :  μ  ≠ 60

Test statistic = t

= (x̅- μ ) / S / √ n

= (63.67-67) / 18.77 / √ 15

= 0.757

Test statistic = t =  0.757

P-value =0.4614

α = 0.01

P-value > α

0.4614 > 0.01

Fail to reject the null hypothesis .

There is not sufficient evidence to conclude that the original claim that the mean of the population of the estimate is 60 seconds equal to it that of a group the students who are reasonably good at estimating one minute.